Resistors in parallel combine through their reciprocals

1. The formula for two resistors in parallel is $\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$.
2. Rearranged for exactly two resistors it becomes $R=\frac{R_1{R}_2}{R_1+R_2}$, which is quicker to key into a calculator.
3. The answer is always less than the smaller of the two resistances, which is the check to run every single time.
4. Two identical resistors give exactly half the value of one, so two $10\Omega$ resistors in parallel give $5.0\Omega$.
5. The reason is physical rather than algebraic: a second branch gives the charge an extra route, so more current flows for the same p.d. and the resistance must be lower.

Working the calculation without losing the reciprocal

1. Work out $\frac{1}{R_1}$ and $\frac{1}{R_2}$ first and add them, remembering that the total you now have is $\frac{1}{R}$ and not $R$.
2. Take the reciprocal of that total at the end, because forgetting this last step is the most common error in the whole topic.
3. Keep the full value on the calculator rather than rounding the reciprocals, since rounding early can shift the final answer noticeably.
4. Give the answer in ohms to two or three significant figures, matching the precision of the values you were given.
5. Once you have the combined resistance you can treat the pair as a single resistor and find the supply current from $I=\frac{V}{R}$.